Solid propellants — fuel + oxidizer in polymer matrix: Worked examples

3.3.35 · D3 · Physics › Rocket Propulsion › Solid propellants — fuel + oxidizer in polymer matrix

Shuru karne se pehle, yeh hain woh sirf wahi tools jo hum use karenge, har ek plain words mein stated taaki koi symbol unearned na lage:

Recall Woh chaar relations jinpar hum tik te hain (parent note se)

Thrust (ideal): $F = \overset{m}{˙} v_{e}$ — force barabar hai kitna mass per second nikal raha hai, times kitni tezi se nikal raha hai.
Thrust (full): $F = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ — nozzle ke mooh par pressure mismatch ka ek push/pull aur add karo.
Mass burn rate: $\overset{m}{˙} = ρ_{p} A_{b} r$ — density $\times$ burning surface area $\times$ kitni tezi se flame andar ki taraf kha rahi hai.
Burn-rate law: $r = a p_{c}^{n}$ — flame speed chamber pressure ke saath badhti hai; stable tabhi jab $n < 1$ .

Yahan $\overset{m}{˙}$ (padho "m-dot") ka matlab hai "kilograms of solid jo gas mein convert ho raha hai per second". Dot shorthand hai "rate of change with time" ke liye.

The scenario matrix

Har solid-motor numerical problem in cells mein se ek hai. Neeche ke examples har cell ke saath tag kiye gaye hain taaki aap dekh sako ki poora board filled hai.

#	Cell class	Kya tricky banaata hai	Example
A	Baseline — plug into $\overset{m}{˙} = ρ_{p} A_{b} r$ , $F = \overset{m}{˙} v_{e}$	kuch nahi; warm-up	Ex 1
B	Pressure exponent scaling ( $n < 1$ )	ratio, exponent, $a$ cancels	Ex 2
C	Degenerate: $A_{b} \to 0$ (burnout)	thrust zero ho jaata hai — limiting case	Ex 3
D	Runaway: $n \geq 1$	unstable feedback — woh forbidden case	Ex 4
E	Geometry: progressive (cylindrical port badhta hai)	$A_{b}$ time ke saath badhta hai	Ex 5
F	Geometry: neutral (star / end-burner)	$A_{b}$ constant rakha jaata hai	Ex 5
G	Pressure term: vacuum vs sea level	sign of $(p_{e} - p_{a}) A_{e}$ flip hota hai	Ex 6
H	Real-world word problem (mission burn time + total impulse)	multi-step, unit-heavy	Ex 7
I	Exam twist ( $I_{s p}$ from $v_{e}$ ; feeds Tsiolkovsky)	teen formulas connect karta hai	Ex 8

Cells A–I sab cover hain Examples 1–8 se.

Statement. Ek motor mein $ρ_{p} = 1750 kg/m^{3}$ hai, burning area $A_{b} = 0.040 m^{2}$ , burn rate $r = 9 mm/s$ , exhaust speed $v_{e} = 2500 m/s$ , aur yeh matched chal raha hai ( $p_{e} = p_{a}$ ). $\overset{m}{˙}$ aur thrust $F$ nikalo.

Forecast: Aage padhne se pehle nearest kN mein thrust guess karo. (Hint: $\overset{m}{˙}$ 1 kg/s se kam hoga, $v_{e}$ hai 2500 → toh $F$ kuch kN hoga.)

Burn rate convert karo: $r = 9 mm/s = 0.009 m/s$ . Yeh step kyun? Baaki har quantity metres aur kilograms mein hai; millimetres mix karne se answer $1000 \times$ galat ho jaayega.
Mass burn rate: $\overset{m}{˙} = ρ_{p} A_{b} r = 1750 \times 0.040 \times 0.009 = 0.63 kg/s$ . Yeh step kyun? Ek second mein flame $0.009$ m peeche jaati hai poore $0.040 m^{2}$ face par, $0.00036 m^{3}$ solid khodkar; times density se kilograms milte hain.
Thrust: $F = \overset{m}{˙} v_{e} = 0.63 \times 2500 = 1575 N$ . Yeh step kyun? Kyunki $p_{e} = p_{a}$ hai toh pressure term exactly zero hai, isliye ideal thrust sirf momentum flux hai.

Verify: Units: $(kg/s) (m/s) = kg\cdotm/s^{2} = N$ ✓. Magnitude: $\approx 1.6$ kN, ek small-motor ki value — sensible. ✓

Example 2 — Pressure exponent scaling (cell B)

Worked example

Statement. Ek propellant $r_{1} = 9 mm/s$ par burn karta hai jab chamber pressure $p_{1} = 6 MPa$ hai. Uska pressure exponent $n = 0.40$ hai. $p_{2} = 9 MPa$ par burn rate kya hoga?

Forecast: Pressure $50%$ badh gaya. Kyunki $n < 1$ hai, kya burn rate $50%$ se zyada ya kam badhega?

Burn-rate law ka ratio likho: $\frac{r _{2}}{r _{1}} = \frac{a p _{2}^{n}}{a p _{1}^{n}} = (\frac{p _{2}}{p _{1}})^{n}$ . Yeh step kyun? Woh messy temperature-dependent constant $a$ dono states mein identical hai, isliye divide karne se woh vanish ho jaata hai. Hume uski value ki kabhi zaroorat hi nahi padti.
Numbers plug karo: $\frac{p _{2}}{p _{1}} = \frac{9}{6} = 1.5$ , toh $\frac{r _{2}}{r _{1}} = 1. 5^{0.40}$ . Yeh step kyun? Power $n = 0.40$ par raise karna yahi dikhata hai ki "pressure change ka kitna fraction flame tak pahunchta hai."
Power evaluate karo: $ln 1.5 = 0.4055$ , $\times 0.40 = 0.1622$ , $e^{0.1622} = 1.176$ . Toh $r_{2} = 9 \times 1.176 = 10.6 mm/s$ . Yeh step kyun? Hum $ln$ phir $e^{(\cdot)}$ use karte hain kyunki yahan koi "nice" root nahi hai — exponential logarithm ko undo karta hai $1. 5^{0.4}$ recover karne ke liye.

Verify: Pressure $50%$ badha lekin burn rate sirf $\approx 18%$ badha — pressure change se kam, exactly wohi jo $n < 1$ guarantee karta hai. Yahi self-limiting behaviour hai jiski wajah se motor runaway nahi karta. ✓

Example 3 — Degenerate case: burnout, $A_{b} \to 0$ (cell C)

Worked example

Statement. Ek end-burner ki face area burn karte waqt constant rehti hai, lekin aakhri instant mein propellant ka aakhri sliver gayab ho jaata hai aur $A_{b} \to 0$ . Example-1 motor lo ( $ρ_{p} = 1750$ , $r = 0.009$ m/s, $v_{e} = 2500$ ) lekin $A_{b} = 0.010 m^{2}$ lo, phir $A_{b} = 0.001 m^{2}$ , phir $A_{b} = 0$ . Har ek se kya thrust milta hai?

Forecast: Jaise $A_{b}$ zero ki taraf shrink karta hai, kya thrust smoothly taper karta hai ya cliff se girta hai?

Area ka function as thrust: $F (A_{b}) = ρ_{p} A_{b} r v_{e} = (1750) (0.009) (2500) A_{b} = 39375 A_{b}$ . Yeh step kyun? Dono relations chain karne se pata chalta hai thrust directly proportional hai burning area se — origin se guzarne wali ek straight line.
Evaluate karo: $A_{b} = 0.010$ par $\Rightarrow F = 393.75 N$ ; $A_{b} = 0.001$ par $\Rightarrow F = 39.375 N$ ; $A_{b} = 0$ par $\Rightarrow F = 0 N$ . Yeh step kyun? Har ek bas $39375$ times area hai; limit $A_{b} \to 0$ se thrust $\to 0$ continuously milta hai.
Limit interpret karo: jis pal burning surface disappear hoti hai, poora thrust bhi chala jaata hai — motor "stored force" par coast nahi kar sakta. Thrust tabhi exist karta hai jab kuch burn ho raha ho. Yeh step kyun? Yeh poore model ki degenerate boundary hai: no $A_{b}$ , no $\overset{m}{˙}$ , no $F$ .

Verify: $F$ linear hai $A_{b}$ mein, isliye area ko half/tenth karne se thrust exactly tenth ho jaata hai ( $393.75 \to 39.375$ ) ✓, aur $F (0) = 0$ bina kisi discontinuity ke ✓. Units check on the slope $ρ_{p} r v_{e}$ : $(kg/m^{3}) (m/s) (m/s) = kg / (m\cdots^{2})$ ; $A_{b}$ ko $m^{2}$ mein multiply karne par $(kg / (m\cdots^{2})) (m^{2}) = kg\cdotm/s^{2} = N$ ✓.

Example 4 — The forbidden case: runaway when $n \geq 1$ (cell D)

Worked example

Statement. Do propellants $p_{1} = 5 MPa$ , $r_{1} = 8 mm/s$ par hain. Ek random flutter pressure ko $20%$ badhakar $p_{2} = 6 MPa$ kar deta hai. Safe propellant ( $n = 0.4$ ) aur forbidden one ( $n = 1.3$ ) ke liye burn-rate response compare karo. Kaun sa feedback loop blow up karta hai?

Forecast: Zyada burn rate → zyada gas → zyada pressure. Yeh loop kab settle hota hai, aur kab explode karta hai?

Safe propellant: $\frac{r _{2}}{r _{1}} = 1. 2^{0.4}$ . $ln 1.2 = 0.1823$ , $\times 0.4 = 0.0729$ , $e^{0.0729} = 1.0757$ . Toh burn rate $\approx 7.6%$ badhta hai $20%$ pressure bump ke liye. Yeh step kyun? Response ( $7.6%$ ) disturbance ( $20%$ ) se chhota hai: extra gas pressure spike ko sustain nahi kar sakta, toh woh relax ho jaata hai. Stable.
Forbidden propellant: $\frac{r _{2}}{r _{1}} = 1. 2^{1.3}$ . $ln 1.2 = 0.1823$ , $\times 1.3 = 0.2370$ , $e^{0.2370} = 1.267$ . Burn rate $\approx 26.7%$ badhta hai. Yeh step kyun? Ab response ( $26.7%$ ) disturbance ( $20%$ ) se bada hai: zyada gas → zyada pressure → aur bhi zyada gas. Loop khud ko amplify karta hai.
Conclude karo: boundary exactly $n = 1$ par hai. $n < 1$ ke liye disturbances shrink hoti hain; $n \geq 1$ ke liye woh unboundedly badhti hain → detonation. Isliye parent note $n < 1$ par insist karta hai. Yeh step kyun? Yeh design rule ko ek vague "keep it low" se alag, ek limiting value se pin karta hai.

Verify: Knife-edge $n = 1$ par: $1. 2^{1} = 1.20$ , yaani burn rate exactly utna hi badhta hai jitna pressure — na damping na amplifying (marginal). Safe side $1.076 < 1.20 < 1.267$ forbidden side ✓, expected $n = 1$ marginal value ko straddle karte hue.

Example 5 — Grain geometry: progressive vs neutral (cells E & F)

Figure 1 — Thrust ratio vs distance burned outward, tube port (progressive) aur star port (neutral) ke liye.

Figure — Solid propellants — fuel + oxidizer in polymer matrix

Intuition

Figure 1 kya dikhata hai. Horizontal axis hai flame port wall se kitni door bahar ki taraf burn ho chuki hai, millimetres mein (0 se 10 mm tak). Vertical axis hai thrust divided by uski starting value — ek pure ratio, toh dono curves 1.00 se start hoti hain. Magenta solid curve ek plain cylindrical (tube) port hai: jaise flame bahar ki taraf khaati hai, port circumference badhta hai, toh burning area — aur thrust — steadily climb karta hai 1.50 tak (ek badhta hua "progressive" curve). Violet dashed line ek star port hai: woh poore time 1.00 par pinned rehti hai (ek flat "neutral" curve). Orange dots mark karte hain start (radius 20 mm, thrust ×1.00) aur tube ka end (radius 30 mm, thrust ×1.50).

Jo takeaway aapko dikhni chahiye: sirf geometry decide karti hai ki thrust badhega ya flat rahega.

Worked example

Statement. Ek cylindrical-port grain ki length $L = 0.5 m$ hai, port radius $R_{0} = 20 mm$ se start karta hai. Burning surface inner wall hai, $A_{b} = 2 π R L$ . Yeh $10 mm$ bahar ki taraf burn karta hai. (a) $A_{b}$ kitne factor se badhta hai (progressive)? (b) Ek star grain iske bajaye $A_{b}$ constant rakhta hai (neutral) — agar woh same $A_{b}$ aur thrust se start kare, toh same moment par uska thrust ratio kya hai? Use karo $ρ_{p} = 1750$ , $r = 0.009$ m/s, $v_{e} = 2500$ .

Forecast: Tube ke liye, thrust port radius follow karta hai. Guess karo percentage thrust rise jab radius 20 se 30 mm ho jaaye.

Lengths metres mein convert karo aur initial area nikalo: $R_{0} = 20 mm = 0.020 m$ , toh $A_{b, 0} = 2 π R_{0} L = 2 π (0.020) (0.5) = 0.0628 m^{2}$ . Yeh step kyun? Ek internal cylinder ki burning surface uski lateral (side) area hoti hai, $2 π R L$ — ek rectangle joh circumference $2 π R$ aur height $L$ wali tube mein roll ho jaata hai. Hum metres mein kaam karte hain toh area m² mein aata hai jo $ρ_{p}$ ke kg/m³ se match kare.
10 mm burn karne ke baad final radius: $R_{1} = 20 + 10 = 30 mm = 0.030 m$ . Final area $A_{b, 1} = 2 π R_{1} L = 2 π (0.030) (0.5) = 0.0942 m^{2}$ . Yeh step kyun? Flame front radially bahar ki taraf khaata hai, toh port radius 10 mm se badhta hai (yahan 0.010 m mein convert kiya) aur burning circumference uske saath badhti hai.
Progressive ratio: $\frac{A _{b, 1}}{A _{b, 0}} = \frac{R _{1}}{R _{0}} = \frac{30}{20} = 1.5$ . Kyunki thrust $\propto A_{b}$ hai, thrust $50%$ badhta hai — ek undesirable rising curve. Yeh step kyun? Ek tube ke liye, $A_{b}$ proportional hai $R$ se, toh area ratio bas radius ratio hai; $L$ , $2 π$ cancel ho jaate hain (aur mm↔m conversion bhi, kyunki dono radii par apply hota hai). Yeh hai magenta curve jo 1.50 tak pahunchti hai.
Neutral (star): design se $A_{b}$ constant rehta hai kyunki extra star points recede karte hain widening ko offset karne ke liye. Thrust ratio $= 1.00$ — ek flat curve. Yeh step kyun? Ek star port fresh burning surface exactly wahan add karta hai jahan ek plain tube use kho deta, $A_{b}$ aur isliye thrust constant rakhta hai — woh violet dashed line 1.00 par held.

Verify: Initial thrust $F_{0} = ρ_{p} A_{b, 0} r v_{e} = 1750 (0.0628) (0.009) (2500) = 2472 N$ . Progressive final $F_{1} = 1.5 F_{0} = 3708 N$ . Ratio $3708/2472 = 1.5$ ✓. Neutral final $= F_{0} = 2472 N$ , ratio $1.00$ ✓.

Example 6 — Pressure term: vacuum vs sea level (cell G)

Worked example

Statement. Ek motor mein $\overset{m}{˙} = 4 kg/s$ , $v_{e} = 2600 m/s$ , nozzle exit area $A_{e} = 0.05 m^{2}$ , aur exit pressure $p_{e} = 70 kPa$ hai. Thrust nikalo (a) sea level par ( $p_{a} = 101.3 kPa$ ) aur (b) vacuum mein ( $p_{a} = 0$ ). (Upar ki definition se yaad karo: $A_{e}$ exit opening ka area hai, $p_{e}$ wahan gas pressure hai, $p_{a}$ woh outside pressure hai jo push back karta hai.)

Forecast: Full formula $(p_{e} - p_{a}) A_{e}$ add karta hai. Kaun sa environment zyada thrust deta hai, aur kya extra term kabhi subtract karta hai?

Momentum thrust (dono mein same): $\overset{m}{˙} v_{e} = 4 \times 2600 = 10400 N$ . Yeh step kyun? Mass flow aur exhaust speed chemistry/nozzle se set hoti hai, outside air se unaffected.
Sea level pressure term: $(p_{e} - p_{a}) A_{e} = (70000 - 101300) (0.05) = (- 31300) (0.05) = - 1565 N$ . Yeh step kyun? Yahan $p_{e} < p_{a}$ hai (over-expanded nozzle): outside air exit plane par jet se zyada zor se push back karta hai, toh term negative hai — yeh thrust churaata hai.
Sea-level total: $F = 10400 - 1565 = 8835 N$ . Yeh step kyun? Full thrust formula momentum thrust plus pressure term hai; negative number add karna same hai subtract karne se, toh atmospheric back-push net force ko ideal $10400 N$ se neeche le aata hai.
Vacuum pressure term: $(70000 - 0) (0.05) = + 3500 N$ , toh $F = 10400 + 3500 = 13900 N$ . Yeh step kyun? Koi ambient air push back nahi karta, toh exit pressure fully outward act karta hai — term positive ho jaata hai, toh wahi motor space mein zyada thrust produce karta hai.

Verify: Vacuum thrust ( $13900$ N) $>$ sea-level thrust ( $8835$ N) ✓ — real rockets mein climb karte waqt thrust badhta hai. Dono ke beech ka gap exactly $p_{a} A_{e} = 101300 \times 0.05 = 5065 N$ hai, aur indeed $8835 + 5065 = 13900$ ✓. Har term ke units: $(Pa) (m^{2}) = (N/m^{2}) (m^{2}) = N$ , momentum term $(kg/s) (m/s) = N$ se match karta hai ✓.

Example 7 — Real-world word problem (cell H)

Worked example

Statement. Ek booster mein $M_{p} = 900 kg$ propellant hai. Yeh constant $\overset{m}{˙} = 6 kg/s$ par burn karta hai, $v_{e} = 2400 m/s$ ke saath (matched). Nikalo: (a) burn time, (b) constant thrust, (c) total impulse $J = F t_{b}$ , aur (d) confirm karo ki $J$ barabar bhi hai $M_{p} v_{e}$ ke.

Forecast: Do alag routes same total impulse dene chahiye. Guess karo ki $J$ $1$ million ke paas hoga ya $2$ million N·s ke.

Burn time: $t_{b} = \frac{M _{p}}{m ˙} = \frac{900}{6} = 150 s$ . Yeh step kyun? Total mass divided by mass-per-second se seconds milte hain — propellant kitni der chalega.
Thrust: $F = \overset{m}{˙} v_{e} = 6 \times 2400 = 14400 N$ . Yeh step kyun? Constant $\overset{m}{˙}$ aur $v_{e}$ ka matlab hai poore burn mein constant thrust.
Total impulse via force×time: $J = F t_{b} = 14400 \times 150 = 2160000 N\cdots$ . Yeh step kyun? Impulse accumulated push hai; constant force ke saath yeh simply force times duration hai.
Cross-check via mass×speed: $J = M_{p} v_{e} = 900 \times 2400 = 2160000 N\cdots$ . Yeh step kyun? Total impulse barabar hai sab momentum jo peeche throw kiya gaya, jo total ejected mass times uski speed hai — ek aise route se jo kabhi time mention nahi karta.

Verify: Dono routes $2.16 \times 1 0^{6} N\cdots$ dete hain ✓. Units: $N\cdots = kg\cdotm/s$ (momentum) ✓. Yeh value Specific Impulse aur Tsiolkovsky Rocket Equation ko feed karta hai delivered $Δ v$ ke liye.

Example 8 — Exam twist: $v_{e}$ se specific impulse aur $Δ v$ (cell I)

Yahan do nayi quantities aati hain; dono ko plain words mein define kiya gaya hai pehle use karne se.

Worked example

Statement. Example-7 booster ke liye ( $v_{e} = 2400 m/s$ , $M_{p} = 900 kg$ ), total vehicle wet mass $m_{0} = 1200 kg$ lo (toh dry mass $m_{f} = 1200 - 900 = 300 kg$ ). Nikalo (a) specific impulse $I_{s p}$ with $g_{0} = 9.81 m/s^{2}$ , aur (b) ideal velocity gain $Δ v$ .

Forecast: $I_{s p}$ couple-hundred-seconds range mein typical hai solids ke liye. $Δ v$ guess karo — kya yeh $v_{e}$ se zyada hoga?

Specific impulse: $I_{s p} = \frac{v _{e}}{g _{0}} = \frac{2400}{9.81} = 244.6 s$ . Yeh step kyun? Exhaust speed ko $g_{0}$ se divide karna "gas kitni tezi se jaata hai" ko us standard efficiency figure mein convert karta hai jo engineers seconds mein quote karte hain — ek hi step chahiye jab $v_{e}$ pata ho.
Mass ratio: $\frac{m _{0}}{m _{f}} = \frac{1200}{300} = 4$ . Yeh step kyun? Rocket equation sirf full to empty mass ke ratio ki parwah karta hai — kitna bada fraction throwable tha — toh hum pehle woh pure number compute karte hain.
Rocket equation: $Δ v = v_{e} ln (m_{0} / m_{f}) = 2400 \times ln 4$ . $ln 4 = 1.3863$ , toh $Δ v = 2400 \times 1.3863 = 3327 m/s$ . Yeh step kyun? Hum natural log use karte hain kyunki velocity fractional mass ke girne ke saath build up hoti hai — har equal fractional mass loss ek equal velocity chunk add karta hai, aur $ln$ exactly "equal fractional steps ka sum" hai.

Verify: $I_{s p} \approx 245 s$ solid-motor range ( $\sim$ 200–290 s) mein squarely hai ✓. Aur $Δ v = 3327 > v_{e} = 2400$ : kyunki mass ratio $e^{1} \approx 2.718$ se exceed karta hai, ship apni khud ki exhaust speed se zyada tez ho jaata hai ✓. $Δ v$ ke units: m/s ( $ln$ dimensionless hai) ✓.

Recall

Recall Kaun sa cell kaun sa tha?

Baseline plug-in ::: Ex 1 (cell A)
$a$ burn-rate ratio mein cancel ho jaata hai ::: Ex 2 (cell B)
Thrust $\to 0$ jaise $A_{b} \to 0$ ::: Ex 3 (cell C)
Runaway boundary $n = 1$ par ::: Ex 4 (cell D)
Tube $A_{b}$ badhata hai (progressive), star flat rakhta hai (neutral) ::: Ex 5 (cells E, F)
Pressure term vacuum vs sea level mein sign flip karta hai ::: Ex 6 (cell G)
Total impulse do tareekon se ::: Ex 7 (cell H)
$I_{s p}$ aur $Δ v$ from $v_{e}$ ::: Ex 8 (cell I)

Recall Teen exit-plane symbols

$A_{e}$ kya hai? ::: Nozzle exit opening ka area (m²).
$p_{e}$ kya hai? ::: Exit plane par exhaust gas pressure (Pa).
$p_{a}$ kya hai? ::: Ambient (surrounding) pressure jo push back karta hai; $\approx 101.3$ kPa sea level par, $0$ vacuum mein.
$I_{s p}$ kya hai aur uska unit kya hai? ::: Specific impulse $= v_{e} / g_{0}$ , seconds mein measure hota hai.

Connections

Parent topic — woh chaar master relations jo yeh examples exercise karte hain.
Specific Impulse — Ex 8 $v_{e}$ ko $I_{s p}$ mein turn karta hai.
Tsiolkovsky Rocket Equation — Ex 7–8 total impulse aur $Δ v$ feed karte hain.
De Laval Nozzle — $v_{e}$ , $p_{e}$ , $A_{e}$ set karta hai jo Ex 6 mein use hue.
Newton's Third Law — har thrust value ke peeche momentum principle.
Liquid Propellants — fixed geometry ke dikhaye throttle/restart limits se contrast.

Worked examples — Solid propellants — fuel + oxidizer in polymer matrix

The scenario matrix

Example 1 — Baseline (cell A)

Example 2 — Pressure exponent scaling (cell B)

Example 3 — Degenerate case: burnout, $A_{b} \to 0$ (cell C)

Example 4 — The forbidden case: runaway when $n \geq 1$ (cell D)

Example 5 — Grain geometry: progressive vs neutral (cells E & F)

Example 6 — Pressure term: vacuum vs sea level (cell G)

Example 7 — Real-world word problem (cell H)

Example 8 — Exam twist: $v_{e}$ se specific impulse aur $Δ v$ (cell I)

Recall

Connections